Question: Factor completely. $49q^4-4=$
Solution: $49 q ^4 -4 = ({7 q ^2})^2-({2 })^2$ Using the difference of squares pattern: $({7 q ^2})^2 - ({2 })^2 =({7 q ^2}+{2 })({7 q ^2}-{2 })$ In conclusion, $49 q ^4-4 =(7 q ^2+2 )(7 q ^2-2 )$ Remember that you can always check your factorization by expanding it.